Posted by Halle on .
Hi there
I need help with the restrictions on the variables of this question:
Simplify. State any restrictions on the variables.
log(x^2+7x+12)/log(x^29)
So my answer is: log(x+4/x3) which is correct.
Now for the restrictions, I have:
x<4 and x>3
However the back of the book says the answer is only x>3 and not x<4 for the restrictions of the variables. But when I sub in any number such as 5 into the ORIGINAL log equation, i get a defined and positive log of x value. So? is mine correct or theirs? Explain fully.

Math 
Halle,
Sorry the question is:
log[(x^2+7x+12)/x^29)] 
Math 
Reiny,
first of all
log(x^2+7x+12)/log(x^29) is NOT equal to
log((x+4)/(x3)) like you say
If the back of your book has that answer, then they are definitely WRONG
log [(x^2+7x+12)/(x^29)]
would be equal to log((x+4)/(x3))
to find the restrictions you have to consider the numerator and denominator
remember we can only take logs of positive numbers
at the top:
log(x^2+7x+12) = log[(x+3)(x+4)]
x < 4 OR x > 3
at the bottom:
log(x^2  9) = log[(x3)(x+3)]
x < 3 or x > 3
but both of these conditions must be met, so
x < 4 OR x > 3 AND x < 3 or x > 3
which leaves you with x < 4 or x > 3
notice that between 4 and +3 we get a negative for either one or the other function
another way to understand the restriction would be:
graph both y = (x+3)(x+4) and
y = (x+3)(x3)
any x where either or both of those graphs dip below the xaxis would not be allowed.
so yes, they are wrong with their restrictions. 
Math 
Reiny,
ARGHHHH! saw your correction only after I posted.
So my answer would match your first posting, and it would clearly be a more interesting question. 
Math 
Reiny,
so now
log [(x+4)(x+3)/((x3)(x+3))]
= log(x+4) + log(x+3)  log(x+3)  log(x3)
now each of those terms must be defined.
so x>4 AND x>3 AND x>3
which is x > 3
notice that we canceled out log(x+3)
but for all x ≤ 3 that term would be undefined. So can we really do algebra with undefined numbers??
I know this is a bit tricky, since 5 (or any x < 4) would give a positive result in the original algebraic expression, but it would not work in the individuals terms.